Find any critical numbers of the function f(Ø)= 2secØ +tanØ, 10π < Ø < 12π

A. 67π/6
B. 71π/6
C. 63π/6
D. A and B

well,

f'(Ø) = 2secØ tanØ + sec^2Ø
= sec^2Ø(2tanØ cosØ + 1)
= sec^2Ø(2sinØ+1)

f'(Ø)=0 when sinØ = -1/2

Now, what else is a critical number?

To find the critical numbers of the function f(Ø) = 2secØ + tanØ, we need to first find the derivative of the function and then solve for when the derivative equals zero.

The derivative of f(Ø) can be found using the chain rule. Let's start by finding the derivative of secØ and tanØ separately.

The derivative of secØ can be found using the quotient rule. Recall that secØ = 1/cosØ.

Derivative of secØ:
d(secØ)/dØ = (d/dØ)(1/cosØ)
= -1/cosØ * d(cosØ)/dØ
= -secØ * sinØ

The derivative of tanØ can be found using the quotient rule as well. Recall that tanØ = sinØ/cosØ.

Derivative of tanØ:
d(tanØ)/dØ = (d/dØ)(sinØ/cosØ)
= (cosØ * d(sinØ)/dØ - sinØ * d(cosØ)/dØ) / cos²Ø
= (cosØ * cosØ - sinØ * (-sinØ)) / cos²Ø
= (cos²Ø + sin²Ø) / cos²Ø
= 1 / cos²Ø

Now, let's find the derivative of the function f(Ø) = 2secØ + tanØ by adding the derivatives of secØ and tanØ.

f'(Ø) = -2secØ * sinØ + 1 / cos²Ø

To find the critical numbers, we set the derivative equal to zero and solve for Ø.

0 = -2secØ * sinØ + 1 / cos²Ø

Simplifying the equation, we get:

2secØ * sinØ = 1 / cos²Ø

2 * sinØ = cosØ / cos²Ø

2sin²Ø = 1

sin²Ø = 1/2

Taking the square root of both sides:

sinØ = ± √(1/2)

We know that the sine function is positive (greater than zero) in the given interval from 10π to 12π. Therefore, we can ignore the negative sign and solve for Ø using sinØ = √(1/2).

sinØ = √(1/2) can be rewritten as:

Ø = arcsin(√(1/2))

Using the unit circle, we know that sin(π/4) = √(1/2). So,

Ø = π/4

However, the given interval is from 10π to 12π, so we need to add 2π to the solution:

Ø = π/4 + 2π = 9π/4

Thus, the critical number of the function in the given interval is Ø = 9π/4.

Now, let's check which answer choice includes the critical number within the given interval.

10π < Ø < 12π

10π < 9π/4 < 12π

40π < 9π < 48π

Based on the given option, A. 67π/6, we can see that it is not within the given interval.

Therefore, the correct answer is:

D. A and B